Exponential-logarithmic equation

Question

I don't know how to solve this equation:$$(1)\quad e^ {-x} = -\ln x$$

$x$ should be the abscissa of the point $P$ where the two functions meet on the plan and $$ P \in f(x) :y=x$$

so $(1)$ should be equal to $$ e^{-x}=x=-\ln x$$

How do I solve this?

Answer 1

$\quad e^ {-x} = -\ln x\iff e^ {-x}+\ln x=0 $

Let $g(x)=e^ {-x}+\ln x$

$g(1)=\dfrac{1}{e}>0$ and $\displaystyle \lim_{x\to 0} g(x)=-\infty$

The Intermediate value theorem guaranties there exists $c\in (0,1)$ such that $g(c)=0 \iff \exists c\in (0,1) $ such that $e^ {-c} = -\ln c$

Answer 2

Your equation can be rearrange to $x=e^{-e^{-x}}$. Now define $f(x)=e^{-x}$ so we are looking for solutions to $x=f(f(x))$. If we have a solution to $x=f(x)$ then this will also be a solution to $x=f(f(x))$ so to obtain a solution it suffice to solve $x=e^{-x}$. This can easily be achieved by the Lambert $W$ function and gives \begin{eqnarray*} x=W(1)=0.56714329 \cdots \end{eqnarray*}

The value above is calculated here https://www.wolframalpha.com/input/?i=productlog(1)

For more details about the Lambert W function see wiki: https://en.wikipedia.org/wiki/Lambert_W_function